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A vertical spring-mass system with lower end of spring is fixed, made to undergo small oscillations. If the spring is stretched by `25cm`, energy stored in the spring is `5J`. Find the mass of the block if it makes `5` oscillations each second. |
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Answer» Correct Answer - `(16)/(10pi^(2)) = 0.16 Kg` `(1)/(2)Kx^(2) = 5` `(1)/(2) k xx ((25)/(100))^(2) = 5` `k = 160` `omega = 10pi = sqrt((K)/(m))` `m = (K)/(100pi^(2))` `m = (160)/(100pi^(2)) = (16)/(10pi^(2)) = 0.16 kg` |
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